Theorem nrmod 3842

Description: Deduce the negation of a restricted "at most one" quantifier. (Contributed by Thierry Arnoux, 13-Jul-2026.)

Hypotheses

Ref	Expression
nrmod.1	⊢ (𝑥 = 𝑋 → (𝜓 ↔ 𝜒))
nrmod.2	⊢ (𝑥 = 𝑌 → (𝜓 ↔ 𝜃))
nrmod.x	⊢ (𝜑 → 𝑋 ∈ 𝐴)
nrmod.y	⊢ (𝜑 → 𝑌 ∈ 𝐴)
nrmod.3	⊢ (𝜑 → 𝜒)
nrmod.4	⊢ (𝜑 → 𝜃)
nrmod.5	⊢ (𝜑 → 𝑋 ≠ 𝑌)

Assertion

Ref	Expression
nrmod	⊢ (𝜑 → ¬ ∃*𝑥 ∈ 𝐴 𝜓)

Distinct variable groups:   𝑥,𝐴   𝑥,𝑋   𝑥,𝑌   𝜒,𝑥   𝜃,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem nrmod

Step	Hyp	Ref	Expression
1		nrmod.5	. . . . 5 ⊢ (𝜑 → 𝑋 ≠ 𝑌)
2	1	neneqd 2962	. . . 4 ⊢ (𝜑 → ¬ 𝑋 = 𝑌)
3		nrmod.y	. . . . 5 ⊢ (𝜑 → 𝑌 ∈ 𝐴)
4		nrmod.4	. . . . 5 ⊢ (𝜑 → 𝜃)
5	3, 4	jca 521	. . . 4 ⊢ (𝜑 → (𝑌 ∈ 𝐴 ∧ 𝜃))
6	2, 5	2thd 268	. . 3 ⊢ (𝜑 → (¬ 𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃)))
7		nbbn 386	. . 3 ⊢ ((¬ 𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃)) ↔ ¬ (𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃)))
8	6, 7	sylib 221	. 2 ⊢ (𝜑 → ¬ (𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃)))
9		nrmod.x	. . . . 5 ⊢ (𝜑 → 𝑋 ∈ 𝐴)
10		nrmod.3	. . . . 5 ⊢ (𝜑 → 𝜒)
11	9, 10	jca 521	. . . 4 ⊢ (𝜑 → (𝑋 ∈ 𝐴 ∧ 𝜒))
12	11	biantrud 541	. . 3 ⊢ (𝜑 → (∃*𝑥 ∈ 𝐴 𝜓 ↔ (∃*𝑥 ∈ 𝐴 𝜓 ∧ (𝑋 ∈ 𝐴 ∧ 𝜒))))
13		nrmod.1	. . . 4 ⊢ (𝑥 = 𝑋 → (𝜓 ↔ 𝜒))
14		nrmod.2	. . . 4 ⊢ (𝑥 = 𝑌 → (𝜓 ↔ 𝜃))
15	13, 14	rmob 3840	. . 3 ⊢ ((∃*𝑥 ∈ 𝐴 𝜓 ∧ (𝑋 ∈ 𝐴 ∧ 𝜒)) → (𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃)))
16	12, 15	biimtrdi 256	. 2 ⊢ (𝜑 → (∃*𝑥 ∈ 𝐴 𝜓 → (𝑋 = 𝑌 ↔ (𝑌 ∈ 𝐴 ∧ 𝜃))))
17	8, 16	mtod 201	1 ⊢ (𝜑 → ¬ ∃*𝑥 ∈ 𝐴 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ≠ wne 2957  ∃*wrmo 3366

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2566  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-rmo 3367  df-v 3455

This theorem is used by:  prlngplngtr  29302